Math of Triangular and Fibonacci Numbers Yahoo Group Recursive Series Problem =============================================== ramsey2879 Message 1 of 4 May 28, 2005 ----------------------------------------------- T(n) = n(n+1)/2 are the well known triangular numbers. Let A(0) = 2 let A(1) = 3 A(n) = 6*A(n-1) -A(n-2) - 4. This series gives 2, 3, 12, 65 etc. The product of any two adjacent terms is always a triangular number. For instance 2*3 = T(3); 3*12 = T(8), 12*65 = T(39) etc. In fact A(0) can be any integer and still there would always be an infinite number of solutions in integers for A(1) and B such that the recursive series defined by A(0), A(1) and A(n) = 6*A(n-1) - A(n-2) - B has the property that the product of two adjacent terms is always a triangular number. The problem is to find at least one solution for A(1) and B for each of A(0) = 1, 4, 5, 6, 7, 8 and 9 respectfully. A solution for A(0) = 2 or 3 is the example given above. Bonus: explain a rule, formula or workable method for finding solutions for any given A(0) that does not simply involve trial and error or computer searching. =============================================== mathiseasyaspi Message 2 of 4 Oct 20, 2005 ----------------------------------------------- --- In Triangular_Numbers@yahoogroups.com, "ramsey2879" wrote: > ... > The problem is to find at least one solution for A(1) and B for each > of A(0) = 1, 4, 5, 6, 7, 8 and 9 respectfully. A solution for A(0) = > 2 or 3 is the example given above. A solution for 1 is A(1) = 6, B = 0. Putting A'(0) =A(1) B=0 gives the same series as a solution for 6, i.e. A'(1) = A(2) = 35. I got this by studying your post on square triangular numbers, can you provide me with the formula or formula's. I have given up. Thanks in advance. =============================================== ramsey2879 Message 3 of 4 Mar 3, 2006 ----------------------------------------------- --- In Triangular_Numbers@yahoogroups.com, "ramsey2879" wrote: > ... > The problem is to find at least one solution for A(1) and B for each > of A(0) = 1, 4, 5, 6, 7, 8 and 9 respectfully. A solution for A(0) = > 2 or 3 is the example given above. This series is listed in the on-line encyclopedia of integer sequences at http://www.research.att.com/~njas/sequences/A108261 =============================================== ramsey2879 Message 4 of 4 Mar 9, 2006 ----------------------------------------------- First I determined that the 2nd order recursive sequence {...a, b, c ...} where a,b,c are any three sucessive terms and c = 6b-a + 2k, has the following property. 8ab - (a+b-k)^2 = 8bc - (b+c-k). That is eight times the product of two adjacent terms always equals the square of the difference between k and the sum of the adjacent terms plus a number that is independent of the selection of the adjacent terms. The proof by induction follows: 8ab - (a+b-k)^2 = 8bc - (b+c-k)^2 ...=8b(6b - a +2k) - (7b - a + k)^2 ...=(48-49)b^2 -a^2 -k^2 +(-8 +14)ab +(16-14)bk +2ak ...=8ab -(b^2 +a^2 +k^2 +2ab -2bk -2ak) ...=8ab -(b+a-k)^2 Thus the above theorm is proven. Now if ab is a triangular number for the first two terms of a sequence, to make the product ab to be a triangular number regardless of which pair of adjacent terms a and b are in the sequence, we need to select a "k" such that 8ab+1=(a+b-k)^2. QED =============================================== Cached by Georg Fischer at Nov 14 2019 12:47 with clean_yahoo.pl V1.4